My experiment is about metals leaching from e-waste (PCB). To know the metals composition of my sample, I did metals digestion that was 1g of sample into 100 ml of aqua regia.
Then, after filtration and dilution, my AAS readings are like 138.8 mg/l, considering the dilution factor of 10. Now, using the common formula of (mg/kg) conversion, I got 13.88 mg/g of Cu which is 1.388% of Cu in our sample.
That's why I'm very confused here because previous studies found a greater percentage of Cu after metals digestion and my case is totally different.
Dear Salehin,
Why you have considered the dilution factor 10? As seen from your question you have digested your 1 g of sample in 100 ml of aqua regia. Have you considered that dilution?
Also you can use online conversion calculators to convert mg/l to % values.
1 mg/l = 0.0001% (i.e. multiply your mg/l value to 0.0001 to get % value).
Regards,
Kunal
Thanks a lot Mr. Kunal. I have considered the dilution factor of 10 because I needed to dilute the filtrate after digestion,, before analysis with AAS.
Thank you Dr. Farooq. I was exactly expecting the result like 13.8%. I again thank you for guiding me with the calculations. Wish you all the best!!!
as you know that:
((milligram / gram) to milligram) / l =
1 kg / m3
this link might be helpful for you..
http://www.endmemo.com/sconvert/mg_kgmg_g.php
as you know that:
((milligram / gram) to milligram) / l =
1 kg / m3
this link might be helpful for you..
http://www.endmemo.com/sconvert/mg_kgmg_g.php
Calculation
Metals in sample = (sample reading-blank reading) x volume (here 100) x dilution factor (here 10)/sample weight
Attached is exact equation when
A: absorbance or concentration from Atomic spect.
0.5 : sample weight
25: dilution of sample
Regards
Hi:
I know this is an old post, but try reading Rüdel, 2007 (Determination of the Elemental Content of Enviromental Samples using ICP-OES). This helped me a lot.
Greetings and good luck.
G. C
=(AAS reading x sample volume up (50 or 100 mL) x Dilution factor)/ Sample weight g
This answer is micro gram/kg unit, you can devide it 1000 and received mg/kg value
I have a confusion in terms of the dilution factor. Please help me out.
I have taken 0.5059g of fortified rice powder and digested with 10ml of mixed (Sulphuric acid and Nitric acid(3:1)). Then after digestion, volume make up to 100ml and AAS reading results is 0.2887ppm or mg/L.
My question is how to calculate in terms of mg/100g of rice.
I tried by following B.K. Kolita Kamal Jinadasa, base on his formula
=(AAS reading x sample volume up (50 or 100 mL) x Dilution factor)/ Sample weight g
so =(0.2887x100x(is it 1?))/05059=57.0666microgram/kg or 0.057066mg/kg
my confusion is what is the dilution factor is case of my procedure?
The answer is about rice is;
[0.2887 ug/mL x 100 mL x 1/1 (dilution factor)]/[0.5059 g] = 57.0666 ug/g
This is mg/100 g = (57.0666 ug/1000 ug/mg) x 100 g
or 5.70666 mg in 100 g
No Kg involved!
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